Question

1. the width of a rectangle is 11 inches less than its length. find the dimensions of the rectangle if he area is 80 square inches. let w = a = lw show set up, then solve. ( )( ) = 0 factor solve answer =

Explanation:

Step1: Define variables

Let $L$ be the length of the rectangle. Then the width $W = L - 11$. The area formula is $A=LW$, and $A = 80$. So we substitute $W$ and $A$ into the formula: $80=L(L - 11)$.

Step2: Expand the equation

Expand $L(L - 11)$ to get $L^{2}-11L=80$. Rearrange it to the standard - quadratic form $L^{2}-11L - 80=0$.

Step3: Factor the quadratic equation

We need to find two numbers that multiply to $- 80$ and add up to $-11$. The numbers are $-16$ and $5$. So we can factor the quadratic as $(L - 16)(L + 5)=0$.

Step4: Solve for $L$

Set each factor equal to zero:
If $L - 16=0$, then $L = 16$.
If $L+5 = 0$, then $L=-5$. But since length cannot be negative, we discard $L=-5$.

Step5: Find the width

Since $W = L - 11$ and $L = 16$, then $W=16 - 11=5$.

Answer:

Length is 16 inches and width is 5 inches.